Given:

AB ∥ CD

∠AGE = 3x

∠GHD = 2x

Concept:

Vertical opposite angles are equal.

If a transversal intersects two lines so that interior angles on the same side of the transversal are supplementary, then the lines are parallel.

Calculation:

∠AGE = ∠BGH = 3x (Vertical opposite angles) 

∠GHD = ∠CHF = 2x (Vertical opposite angles) 

∠BGH + ∠GHD = 180° 

(if a transversal intersects two lines so that interior angles on the same side of the transversal are supplementary, then the lines are parallel.)

⇒ 3x + 2x = 180° 

⇒ 5x = 180° 

⇒ x = 36° 

∴ The value of x is 36° 

The correct option is 2 i.e. 36°

Calculation

Draw a line XB parallel to PA and QC

∠ PAB and ∠ ABX are co interior angles

Let ∠ ABX be x and ∠ XBC be y

⇒ 135° + ∠ x = 180

⇒ ∠ x = 45° 

∠ XBC and ∠ BCQ are co interior angles

⇒ 150° + ∠ y = 180

⇒ ∠ y = 30° 

∠ B = ∠ x + ∠ y 

∠ B = 30°  + 45°  = 75° 

Given:

∠ABC = 80°, ∠BAC = 40° and ∠CDE = 70°

Concept used:

The sum of all the angles of the triangle is 180° 

Vertically opposite angle is equal

Sum of all the angles on the straight line is 180° 

Calculation:

In ΔABC

∠ABC + ∠BAC + ∠ACB = 180° 

⇒ 80° + 40° + ∠ACB = 180° 

⇒ ∠ACB = 180° - 120° 

⇒ ∠ACB = 60°

∠ACB = 60° =  ∠DCE

Now, In ΔCDE

∠DCE + ∠CDE + ∠CED = 180° 

⇒ 60° + 70° + ∠CED = 180° 

⇒ ∠CED = 50° 

Now, ∠CED + ∠DEF = 180°     ----(Linear pair)

⇒ 50° + x = 180° 

⇒ x = 130° 

Given:

The ratio of two complementary angles is 5 : 4

Concept Used:

The sum of the two complementary angles are 90° 

Calculation:

Let two angles be 5x and 4x

5x + 4x = 90

⇒ x = 10

two angles are = 50° and 40° 

Difference between them = 50° - 40° 

⇒ 10° 

∴ required difference is 10° 

Given:

∠AOD = 70°

Calculation:

∠AOD + ∠BOD = 180° (Linear pairs of angles)

⇒ 70° + ∠BOD = 180°

⇒ ∠BOD = 180° - 70°

⇒ ∠BOD = 110°

∴ Supplementary angle of ∠BOD = 180° - 110° = 70°

The correct option is 1 i.e. 70°

Supplementary angle: Two angles are called supplementary when their measures add up to 180°.

Linear pairs of angles: A linear pair is a pair of adjacent angles that makes a straight line.

Given:

In ΔABC, 2∠A = 3∠B = 6∠C

Concept Used:

The sum of all the angles of a triangle is 180°

Calculation:

Let, 2∠A = 3∠B = 6∠C = 6k 

so,  ∠A = 3k,  ∠B = 2k and  ∠C = k

Now, in ΔABC

∠A + ∠B + ∠C = 180°

3k + 2k + k = 180°

6k = 180°

k = 30° 

∠A = 3k = 90° 

∴ ∠A is 90°

Given:

The ratio of the sides of the triangle is 3 ∶ $1\frac{1}{4}$ ∶ $3\frac{1}{4}$

Concept used:

If a² + b² > c², then it is acute triangle

If a² + b² = c², then it is right triangle

If a² + b² < c², then it is obtuse triangle

Calculation:

Let the side of the triangle be 3x, 5x/4 and 13x/4

Now, (3x)² + (5x/4)²

⇒ 9x² + 25x²/16

⇒ 169x²/16

⇒ (13x/4)²

Here, (3x)2 + (5x/4)2 = (13x/4)2

∴ It is the right-angled triangle

Formula used - 

In a regular Polygon,

each interior angle + each exterior angle = 180° 

∠ I + ∠ E = 180° 

Sides = 360°/(exterior angle)

Given - 

internal angle of a polygon = 160° 

Solution - 

∠ I = 160° 

⇒ ∠ E = 20° 

⇒ sides = 360°/20° 

⇒ sides = 18

∴ sides of polygon = 18.

Alternate MethodInterior angle of a regular polygon = $\frac{180\times (n-2)}{n}$

Where

n = number of sides

160 = $\frac{180\times (n-2)}{n}$

160n = 180n - 360

20n = 360 

n = 18

∴ Number of sides of a regular polygon = 18

Given:

Number of diagonals in a polygon is 135

Concept used:

Number of diagonals = n (n – 3)/2

Where n is the number of sides in a polygon

Calculation:

Total number of diagonals is 135

∴ 135 = n (n - 3)/2

⇒ n² – 3n – 270 = 0

⇒ n² – 18n + 15n – 270 = 0

⇒ (n – 18) (n + 15) = 0

⇒ n = 18, -15

Given

AOB is a straight line.

Formula:

If AOB is a straight line, then ∠AOB = 180°.

Calculation:

∠AOD + ∠DOC + ∠COB = 180°

⇒ ∠AOD + 70° + ∠COB = 180°

⇒ ∠AOD + ∠COB = 180° – 70°

⇒ ∠AOD + ∠COB = 110°

⇒ θ + 15° + θ – 5° = 110°

⇒ 2θ + 10° = 110°

⇒ 2θ = 110° – 10°

⇒ 2θ = 100°

⇒ θ = 100°/2

∴ θ = 50°